What is the minimum number of different numerals needed to fill a $3\times 3$ square matrix?

We need to determine the minimum number of different numerals required to fill a $3 \times 3$ matrix such that no two horizontally, vertically, or diagonally adjacent cells contain the same numeral.

Consider the $3 \times 3$ grid:

$\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}$

Observe any $2 \times 2$ submatrix, for example, $\begin{pmatrix} 1 & 2 \\ 4 & 5 \end{pmatrix}.$

Each pair of cells in this block is adjacent (either horizontally, vertically, or diagonally). Hence, all four cells in any $2 \times 2$ block must contain different numerals.

Therefore, at least 4 different numerals are necessary.

Consider the following arrangement using four numerals: $\begin{pmatrix} 4 & 2 & 4 \\ 3 & 1 & 3 \\ 4 & 2 & 4 \end{pmatrix}.$

Verification:

• No two horizontally adjacent cells contain the same numeral.
• No two vertically adjacent cells contain the same numeral.
• No two diagonally adjacent cells contain the same numeral.

Although the numerals $2$, $3$, and $4$ are repeated, identical numerals are always separated by at least one cell and therefore are not adjacent.

Since 4 numerals are necessary and 4 numerals are sufficient, the minimum number of different numerals required is 4.